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Identical springs of steel and copper are equally stretched. On which more work will have to be done? |
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Answer» Solution :Work done in STRETCHING a spring, `W=1/2 F xx DELTAL` where `Delta l` increases in length DUE to applying force, ` therefore W prop Delta l[ because (1)/(2) , F` CONSTANT] `therefore W prop (l)/(Y) [ because Delta l = (Fl )/(AY), F, l, A` are constant] `therefore (W_("steel "))/(W_("copper")) = (Y _("copper "))/( Y _(steel"))` but ` Y _("steel " ) lt W _("copper")` `therefore ` Hence, more work will be done in CASE in spring made of copper. |
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